I answered a question on SO today, however it did't receive any up-votes, so I want to know what's wrong with my code.
The requirement is to compare the input numbers(granted to be positive) without logical, relational or bitwise operators, and insert the corresponding comparison sign between them. For example:
input: 4 6
output: 4 < 6
input: 10 2
output: 10 > 2
input: 2 2
output: 2 = 2
Here is my posted code:
#include <stdio.h>
int main(void)
{
unsigned a, b;
scanf("%u %u", &a, &b);
printf("%u ", a);
char relationship[] = {'<', '=', '>'};
putchar(relationship[!!(a/b) - !!(b/a) + 1]);
printf(" %u", b);
return 0;
}
That OP wants to know how to insert a comparison sign(<
, >
, =
) between the two numbers. But my emphasis is on how to determine the relationship of the two numbers.
See: https://stackoverflow.com/q/35532123/5399734 for the original question.
1 Answer 1
That seems a fine solution. Some minor suggestions:
- Instead of spelling out
relationship
as{'<', '=', '>'}
, you could use simply"<=>"
- Instead of printing
a
, the relation andb
in separate statements, it will be more readable to print in a singleprintf
- The expression
!!(a/b) - !!(b/a)
evaluates to -1, 0, 1, commonly returned by acmp
function. If you give the expression a name by putting it in a variable, that will ring a bell with many readers and make it easier to understand.
Something like this:
unsigned a, b;
scanf("%u %u", &a, &b);
int cmp = !!(a/b) - !!(b/a);
char relation = "<=>"[cmp + 1];
printf("%u %c %u\n", a, relation, b);
-
1\$\begingroup\$
!
is a logical operator per the C spec. \$\endgroup\$chux– chux2016年02月21日 23:53:16 +00:00Commented Feb 21, 2016 at 23:53 -
\$\begingroup\$ He's right, and perhaps we can use (_Bool). Please take a look at my updated code on SO: stackoverflow.com/a/35533732/5399734 for details \$\endgroup\$nalzok– nalzok2016年02月21日 23:58:45 +00:00Commented Feb 21, 2016 at 23:58
!
considered a logical operator? \$\endgroup\$!
operators. See my posted code on SO: stackoverflow.com/a/35533732/5399734 \$\endgroup\$